If R_1\subset R_2 are two real closed fields (R_2 is an extension of R_1), then is it always the case that R_1 contains {R_2}_alg; By the latter I mean algebraic elements of R_2.
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What do you mean by "algebraic elements of $R_2$"? Do you mean those elements that are algebraic over $R_1$? Then the answer is yes. There is a straightforward proof: the only algebraic extension of $R_1$ is $R_1[i]$, and $i \notin R_2$. (where $i^2 = 1$) 


I think I've found it out. If R_1\subset R_2 are two real closed fields, then (R_1)_alg should be the same as (R_2)_alg, since otherwise (R_2)_alg would be a proper real extension of (R_1)_alg contradicting realclosedness of (R_1)_alg. 

